Came up with two proofs for: n = 3.One proof involved a very innovative method using irrational numbers. Unfortunately, Euler made a in his proof.

• Andrew Wiles Fermat

Despite this, his method revealed a very promising approach to Fermat's Last Theorem which was later taken up by,. I discuss the details of this method and Euler's mistake in another.The other proof is less generalizable but still brilliant. This is the proof that I will present in today's blog.The details of this proof are based largely on the work by in his book:.Theorem: Euler's Proof for FLT: n = 3x 3 + y 3 = z 3 has integer solutions - xyz = 0(1) Let's assume that we have solutions x,y,z to the above equation.(2) We can assume that x,y,z are. See for the proof(3) First, we observe that there must exist p,q such that (see for proof):(a) gcd(p,q)=1(b) p,q have opposite parities (one is odd; one is even)(c) p,q are positive.(d) 2p.(p 2 + 3q 2) is a cube.(4) Second, we know that gcd(2p,p 2+3q 2) is either 1 or 3. (see for proof).(5) If gcd(2p,p 2+3q 2)=1, then there must be a smaller solution to Fermat's Last Theorem n=3. (see for proof).(6) Likewise, if gcd(2p,p 2+3q 2)=3, then there must be a smaller solution to Fermat's Last Theorem n=3. (see for proof).(7) But then there is necessarily a smaller solution and we could use the same argument on this smaller solution to show the existence of an even smaller solution.

We have thus shown a condition of.QED. Said.Hi Larry, Could you please teach me how to solve FLT for exponent three according to the information below?

Prove Fermat's Last theorem for n=3: X^3 + Y^3 = Z^3 where X, Y, Z are rational integers, then X, Y, or Z is 0. Hint:. Show that if X^3 + Y^3 = Epsilon. Z^3, where X, Y, Z are quadratic integers in Qsqrt(-3), and epsilon is a unit in Qsqrt(-3), then X, Y, or Z is 0.

(recall how to find all solutions to X^2 + Y^2 =Z^2 since it similar.).Begin with if X^3 + Y^3 = Epsilon. Z^3, where X, Y, Z are quadratic integers in Qsqrt(-3), where epsilon is a unit, then lambda divides X, Y or Z, where lambda is (3+sqrt(-3))/2. Also show that (lambda)^2 is an associate of 3.It will be useful to show that if X is congruent to 1 mod lambda, then X^3 is congruent to 1 (mod lambda)^4). Work out a similar desciption for when X is congruent to -1 mod lambda. Use these fact to show that if X^3 + Y^3 = Epsilon. Z^3, if X and Y are not multiples of lambda, but Z is, then Z is a multiple of (lambda)^2.

Fermat’s Last Theorem (FLT) is perhaps the most famous mathematical theorem of all time. Pierre de Fermat wrote in the margin of a book in 1637 that he had found a marvelous proof that there are no positive whole number solutions to the.

Do this by reducing (mod lamda)^4. Note that X^3 + Y^3 = Epsilon. Z^3 can be factored: (x+y)(x+wy)(x+w^2.y)= Epsilon. Z^3, where w is an appropriately chosen quadratic integer. Consider each of these factors as quadratic integers p, q, r. Express x and y in terms of p, q, and r. The fact that we have three equations and two unknowns indicates there will be some extra constraint on p, q, and r.

Consider how many time lambda occurs in the prime factorization of p, q, and r. Use unique factorization of Qsqrt(-3) to show that except for the factors of lambda that you computed, p, q, and r are cubes times units. Use the extra constraint on p, q, and r to find another solution X^3 + Y^3 = Epsilon. Z^3, where Z has one less factor of lambda. Note that it may be necessary to exchange the roles of x, y, z, -x, -y, or -z. Derive a contradiction along the lines of Fermat's method of descent. Thank you very much.

Andrew Wiles Fermat

Said.No, I don't. I will go through your link. I have more questions which related to part of the FLT proof. I can't do them too.

If I finish proving them, it may help me to figure out more for the proof of FLT for n=3. I have some idea about how to prove FLT for n=3, but I must solve the below question first, then can link up the idea. I need your help again.

I type it here. Could you please teach me how to solve it? Prove that if x^3 + y^3 = z^3, and x, y, z are quadratic integers in Qsqrt(-3), then alpha = (3+sqrt(-3))/2 is belonging to Qsqrt(3) must divide one of x, y, or z. Hint: Reduce the equation modulo alpha^3. Thank you very much. Said.Hi Larry,It's been 3 years since I made comments about Fermat might have it right about his comments.I can prove it but that didn't mean it would be the same as Fermat had thought.

Everybody were trying to find the root. But if the equation is invalid when tripletts are integers to begin with, then what is the point for looking for the root.This year will be 375 years since he conjected. This number is nice be cause it is 3.5^3. I think it's time to reveal the riddle.I believe that with your ability you probably could come up with a proof to FLT. Think outside the box: FLT is a mathematical false statement with integers.Cheers,Joe.

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Said.Hi LarryI need help with a problem. Knowing the result for FLT for n=3 I need to prove that if a positive integer n is divisible by 3, then there are no x.y, z such that x^n+y^n=z^n.What is did so far was say that if 3/n then 3=k.n, for some k positive integer. Then if (x^3, y^3,z^3) is a solution for the exponent k(x^3)^k+(y^3)^k=(z^3)^kand because 3.k=k.3 in integers(x^k)^3+(y^k)^3= (z^k)^3where x^k=a,y^k,=b z^k=c are positive integers. Then there are no, a,b,c positive integers such that a^3+b^3=Z^3.Could that be remotely correct?

Said.Hi there- The PROOF OF PROOFS for n=3 - FERMAT-MURGU QUADRUPLETSARE A MATH BEAUTY IN CONCURRENCE NOW WITH PYTHAGOREAN TRIPLETS.IT A REAL HUNTING FOR AND CONTAIN A COMPLETE MODULAR METHOD BY DEFINITION.BUT ANYWAY WE SOLVED FERMAT'S LAST THEOREM WITH ABSOLUTE ACCURACY AND FORALL n VIA Fermat-Murgu Impossible Equations1. SENT all Fermat Equations Solutions in Irrational Field, without any doubts,even for Z=Integers X,Y must to be Irrational.2. Fermat-Murgu n Media Impassable for Fermat Triplets.We Get it by Analyzing with Ion Murgu Math Millennium Equations all n neighborsaround of a supposed by absurd solutions (X,Y,Z).

(Submitted on 3 Dec 2014) Abstract: Conjecture: The quadratic Diophantine equation system $q^2 = a^2 alpha-b^2 beta-c^2 gamma,$ $pq = (ad)^2 alpha-(be)^2 beta-(cf)^2 gamma,$ $p^2 = (ad^2)^2 alpha-(be^2)^2 beta-(cf^2)^2 gamma,$ has no nontrivial integer solution for $d neq e neq f neq 0$, if there is $ alpha= beta= gamma=1$, or $ alpha neq beta neq gamma neq 0$ with $ alpha a,; beta b,; gamma c;$ and $ alpha neq a,; beta neq b,; gamma neq c$. Fermat's Last Theorem states that the Diophantine equation $x^n+y^n=z^n$ has no nontrivial integer solution (x,y,z) for all n2. A proof was published in 1995 by Andrew Wiles. It is shown that a proof of the above conjecture proves Fermat's Last Theorem for all n5.